Hatcher at page 229 proposes to prove the Borsuk-Ulam theorem using the fact that any continuous map $f \colon\mathbb R P^n \to \mathbb RP^m$, $n > m$, induces the trivial map in cohomology with coefficients in $\mathbb Z / 2 \mathbb Z$. I showed this fact but I can't apply it to finish the proof of the Borsuk-Ulam theorem.
Suppose that there is a map $f \colon S^n \to \mathbb R^n$ such that $f(x) \neq f(-x)$ for any $x$. We can define $g \colon S^n \to S^{n-1}$ by
$$
g(x) = \frac{f(x)-f(-x)}{|f(x)-f(-x)|}.
$$
Then since $g(x) = -g(-x)$ this map descends to the map $\tilde g \colon \mathbb RP^n \to \mathbb RP^{n-1}$. We know that any such map induces a trivial map in cohomology with $\mathbb Z / 2 \mathbb Z$ coefficients. But how to get a contradiction?
Best Answer
You already concluded from the assumption that $\tilde g \colon \mathbb RP^n \to \mathbb RP^{n-1}$ is trivial in cohomology with $\Bbb Z_2$ coefficients. Let us now prove that it induces an isomorphism, which will give the desired contradiction.
Note that $\Bbb Z_2$ is a field, hence $H^1(\Bbb RP^k; \Bbb Z_2)\cong \operatorname{hom}(\pi_1(\Bbb RP^k), \Bbb Z_2)$ by the Universal Coefficient Theorem and the Hurewicz Theorem (since $\pi_1(\Bbb RP^k)= \Bbb Z_2$ is abelian).
Under this identification, the map $H^1(\tilde{g})$ becomes $-\circ \pi_1(\tilde g)$, precomposition with $\pi_1(\tilde g)$. In particular, $H^1(\tilde g)$ is an isomorphism if and only if $\pi_1(\tilde g)$ is one. Let us prove that this is the case by a geometric argument.
Let $\gamma:[0,1]\to \Bbb RP^n$ be a generating loop of $\pi_1(\Bbb RP^n)$. It lifts to a path $\tilde{\gamma}:[0,1]\to S^n$ which joins two antipodal points. Note that $g$ preserves antipodes, hence $g\circ \tilde{\gamma}:[0,1]\to S^{n-1}$ is a map joining two antipodes. Therefore, $g\circ \tilde{\gamma}$ descends to a generating loop in $\pi_1(\Bbb RP^{n-1})$ and by commutativity of the lifting diagram, this loop is precisely $\tilde{g}\circ \gamma = \pi_1(\tilde g)(\gamma)$. Thus, we conclude that $\pi_1(\tilde g)$ is an isomorphism.